Question

prove that 1/sec A-tanA - 1/ cosA = 1/secA - 1/ secA- tanA

Answer 1

Let 'A' be assumed as Alpha as no capital letters are available on Brainly Calculator.

$lhs = \frac{1}{ \sec( \alpha ) } - \tan( \alpha ) - \frac{1 }{ \cos( \alpha ) }$
$= \frac{1}{ \sec( \alpha ) } - \tan( \alpha ) - \sec( \alpha )$
$= ( \frac{1}{ \sec( \alpha ) } - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } ) - \sec( \alpha )$
After this kindly refer the attached where I attached above.

From the first attachment, the answer is
- tan(alpha). ======> 1

RHS:

$= \frac{1 }{ \sec( \alpha ) } - \frac{1}{ \sec( \alpha ) } - tan( \alpha )$

As (1/sec(alpha)) - (1/sec(alpha)) gets zero, the remaining is
- tan(alpha) ========>2

From 1 and 2
LHS. =. RHS
Hence the answer.

Note: There are many ways to solve like this equation. If you don't know further what to do in LHS, better simplify it as much as possible and at the same way you simplify RHS until you get the LHS.
If LHS is equal, RHS should ne equal.
Whether Theta or Alpha or Beta the value would not going to change.